What are the solutions to the equation 0=-16x^2+20x ?

The solutions to the equation 0=-16x^2+20x are x=0,x= 5/4

Find the zeros of 0=-16x^2+20x

The zeros of 0=-16x^2+20x are x=0,x= 5/4

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0 = - 16 x^{2} + 20 x

x = 0, x = \frac{5}{4}

Solution steps

0 = - 16 x^{2} + 20 x

Switch sides

- 16 x^{2} + 20 x = 0

Solve with the quadratic formula

x_{1, 2} = \frac{- 20 \pm 2 0 ^{2} - 4 ( - 16 ) \cdot 0}{2 ( - 16 )}

2 0^{2} - 4 (- 16) \cdot 0 = 20

x_{1, 2} = \frac{- 20 \pm 20}{2 ( - 16 )}

Separate the solutions

x_{1} = \frac{- 20 + 20}{2 ( - 16 )}, x_{2} = \frac{- 20 - 20}{2 ( - 16 )}

x = \frac{- 20 + 20}{2 ( - 16 )} : 0

x = \frac{- 20 - 20}{2 ( - 16 )} : \frac{5}{4}

The solutions to the quadratic equation are:

x = 0, x = \frac{5}{4}

x + 8 + ∣ 4 - x ∣ = 0

(x + 3)^{2} - 3 = 0

c^{3} + 125

\frac{3}{4} = \frac{1}{2} + \frac{d}{5}

7 m + 1 > 15

What are the solutions to the equation 0=-16x^2+20x ?
The solutions to the equation 0=-16x^2+20x are x=0,x= 5/4
Find the zeros of 0=-16x^2+20x
The zeros of 0=-16x^2+20x are x=0,x= 5/4